and define $ q(x) = p(x+1) $. What is the coefficient of $ x^2 $ in $ q(x) $?

February 22, 2026

Understanding $ q(x) = p(x+1) $: Definition, Expansion, and Finding the Coefficient of $ x^2 $

If you’ve encountered a function defined as $ q(x) = p(x+1) $, you’re exploring a fundamental concept in polynomial algebra known as shifting or translation. This transformation shifts the graph of the original polynomial $ p(x) $ horizontally and plays a crucial role in many areas of mathematics, including calculus, differential equations, and approximation theory.

What is $ q(x) = p(x+1) $?

The expression $ q(x) = p(x+1) $ defines a new function $ q $ in terms of the original polynomial $ p $, shifted one unit to the left. To understand this, consider shifting the graph of $ p $ by 1 unit left: each input $ x $ of $ q $ corresponds to $ x+1 $ in $ p $. This transformation does not alter the degree of $ p $ — if $ p $ is a degree-2 polynomial, $ q $ remains degree 2 — but it changes the coefficients through what’s known as polynomial substitution.

Expanding $ q(x) = p(x+1) $: General Approach

Let’s assume $ p(x) $ is a general quadratic polynomial:

$$ p(x) = ax^2 + bx + c $$

Now compute $ q(x) = p(x+1) $ by substituting $ x+1 $ into $ p $:

$$ q(x) = p(x+1) = a(x+1)^2 + b(x+1) + c $$

Expand each term:

$ a(x+1)^2 = a(x^2 + 2x + 1) = ax^2 + 2ax + a $
$ b(x+1) = bx + b $
Constant $ c $

Now combine all terms:

$$ q(x) = ax^2 + 2ax + a + bx + b + c = ax^2 + (2a + b)x + (a + b + c) $$

Coefficient of $ x^2 $ in $ q(x) $

From the expansion:

The coefficient of $ x^2 $ is $ a $, which is the same as the coefficient of $ x^2 $ in $ p(x) $.

This confirms a key fact: when shifting a polynomial $ p(x) $ by replacing $ x $ with $ x+1 $, the leading coefficient (i.e., the $ x^2 $ term) remains unchanged.

Why This Matters

Polynomial Transformations: This shift is widely used to simplify functions, move roots, or analyze local behavior.
Derivative Political Sensitivity: Shifting doesn’t affect the derivative’s form — since $ q'(x) = p'(x+1) $, the derivative is just a shifted version.
Computational Efficiency: In symbolic algebra systems, computing $ p(x+1) $ avoids full expansion but follows predictable coefficient rules.

Summary

Given $ q(x) = p(x+1) $, and with $ p(x) = ax^2 + bx + c $:

$$ q(x) = ax^2 + (2a + b)x + (a + b + c) $$

Thus, the coefficient of $ x^2 $ in $ q(x) $ is $ a $, which equals the coefficient of $ x^2 $ in $ p(x) $. This elegant result reflects how horizontal shifts preserve polynomial degree and leading coefficient.

Understanding $ q(x) = p(x+1) $ deepens insight into function transformations and is essential for advanced study in algebra, calculus, and mathematical modeling.